# How do you list all possible roots and find all factors and zeroes of 5x^3+29x^2+19x-5?

Jun 23, 2016

$5 {x}^{3} + 29 {x}^{2} + 19 x - 5 = \left(x + 1\right) \left(5 x - 1\right) \left(x + 5\right)$

with zeros $x = - 1$, $x = \frac{1}{5}$ and $x = - 5$

#### Explanation:

$f \left(x\right) = 5 {x}^{3} + 29 {x}^{2} + 19 x - 5$

Note that if you reverse the signs on the terms of odd degree then the sum of the coefficients is $0$.

That is: $- 5 + 29 - 19 - 5 = 0$

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Hence $f \left(- 1\right) = 0$ and $\left(x + 1\right)$ is a factor:

$5 {x}^{3} + 29 {x}^{2} + 19 x - 5 = \left(x + 1\right) \left(5 {x}^{2} + 24 x - 5\right)$

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To factor the remaining quadratic use an AC method:

Look for a pair of factors of $A C = 5 \cdot 5 = 25$ which differ by $B = 24$.

The pair $25 , 1$ works, in that $25 \times 1 = 25$ and $25 - 1 = 24$.

Use this pair to split the middle term and factor by grouping:

$5 {x}^{2} + 24 x - 5$

$= 5 {x}^{2} + 25 x - x - 5$

$= \left(5 {x}^{2} + 25 x\right) - \left(x + 5\right)$

$= 5 x \left(x + 5\right) - 1 \left(x + 5\right)$

$= \left(5 x - 1\right) \left(x + 5\right)$

Hence zeros $x = \frac{1}{5}$ and $x = - 5$